Answer
$ x_{1}=\frac{5s+4}{6(s^{2}-3)}$
$ x_{2}=\frac{-4s-15}{4(s^{2}-3)}$
Work Step by Step
Writing the equation in matrix form $A\textbf{x}=\textbf{b} $
$\begin{bmatrix}6s&4\\9&2s\end{bmatrix} \,\,\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix} =\begin{bmatrix}5\\-2\end{bmatrix}$
To have a unique solution, $\textbf{A}$ must be invertible. This happens if and only if $\textbf{det A} \ne0$
$\textbf{det }\begin{bmatrix}6s&4\\9&2s\end{bmatrix}=12s^{2}-36$
To find the limits on s;
$12s^{2}-36=0$
$12s^{2}=36$
$s^{2}=3$
$s=+\sqrt 3, or\,-\sqrt 3$
Therefore for a unique solution, $s\ne+\sqrt 3, or\,s\ne-\sqrt 3$
Compute:
$A_{1}\textbf{(b)}=\begin{bmatrix}5&4\\-2&2s\end{bmatrix} ,\, and\,, det(A_{1}\textbf{(b)})=10s+8$
$A_{2}\textbf{(b)}=\begin{bmatrix}6s&5\\9&-2\end{bmatrix} ,\, and\,, det(A_{2}\textbf{(b)})=12s-45$
By using Cramer's rule, if $s\ne+\sqrt 3, or\,s\ne-\sqrt 3$ that is, A is Invertible, then
$ x_{1}=\frac{det(A_{1}(\textbf{b}))}{det(A_{})}=\frac{10s+8}{12(s^{2}-3)}=\frac{5s+4}{6(s^{2}-3)}$
$ x_{2}=\frac{det(A_{2}(\textbf{b}))}{det(A_{})}=\frac{-12s-45}{12(s^{2}-3)}=\frac{-4s-15}{4(s^{2}-3)}$